Case 1: \(\frac{\theta _p}{2\pi }\)  is irrational.
Write
\({\Re e\,}(a(p^{\nu })e^{-i\phi })=(\sin \theta _p)^{-1}\sin \left(2\pi \left<\!\frac{(\nu +1)\theta _p}{2\pi }\!\right>\right)\cos \left(\frac{\pi j\nu }{r_{\varepsilon }}-\phi \right),\) 
where \(\left<\!\frac{(\nu +1)\theta _p}{2\pi }\!\right>\)  denotes the fractional part of \(\frac{(\nu +1)\theta _p}{2\pi }\) . Note that the sequence \(\left\lbrace \cos \left(\frac{\pi j\nu }{r_{\varepsilon }}-\phi \right)\right\rbrace _{\nu \in \mathbb {N}}\)  is \(t_{\varepsilon }\) -periodic and takes only finitely many different values, with \(t_{\varepsilon }=2r^{\prime }_{\varepsilon }\)  if \(\frac{j}{(r_{\varepsilon },j)}\)  is odd, and \(t_{\varepsilon }=r^{\prime }_{\varepsilon }\)  if \(\frac{j}{(r_{\varepsilon },j)}\)  is even, where \(r^{\prime }_\varepsilon =\frac{r_{\varepsilon }}{(r_{\varepsilon },j)}\) . So, one may split the total range for \(\nu \)  into different arithmetic progressions \(d\pmod { t_{\varepsilon }}\)  where \(1\le d \le t_{\varepsilon }\) , so that when \(\nu \)  runs through each of these arithmetic progressions the \(\cos \) -factor becomes constant and takes always the same value, say \(c_d\) . Accordingly we may write
\(\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })>0\rbrace =\sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d>0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(2\pi \left<\frac{\nu \theta _p}{2\pi }\right>\right)>0}}1+ \sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d<0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(2\pi \left<\frac{\nu \theta _p}{2\pi }\right>\right)<0}}1,\) 
and
\(\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })<0\rbrace =\sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d>0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(2\pi \left<\frac{\nu \theta _p}{2\pi }\right>\right)<0}}1+ \sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d<0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(2\pi \left<\frac{\nu \theta _p}{2\pi }\right>\right)>0}}1.\) 
On the other hand we have
\(\lim _{x\rightarrow \infty }\frac{1}{x}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(2\pi \left<\frac{\nu \theta _p}{2\pi }\right>\right)>0}}1 &=& \lim _{x\rightarrow \infty }\frac{1}{x}\displaystyle \sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(2\pi \left<\frac{\nu \theta _p}{2\pi }\right>\right)\in [0,1]}} 1,\\&=& \lim _{x\rightarrow \infty }\frac{1}{x}\displaystyle \sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{2\pi \left<\frac{\nu \theta _p}{2\pi }\right>\in \left[0,\pi \right]}} 1,\\&=& \frac{1}{2t_\varepsilon },\) 
where we used the fact that the sequence \(\left\lbrace \left<\frac{\nu \theta _p}{2\pi }\right>\right\rbrace _{\nu \in \mathbb {N}}\)  is uniformly distributed \(\pmod {1}\)  in \([0,1]\)  in the last step, (Weyl's equidistribution theorem, see, e.g., [1]}[Example 2.1, pp.8]). Similarly we have
\(\lim _{x\rightarrow \infty }\frac{1}{x}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(2\pi \left<\frac{\nu \theta _p}{2\pi }\right>\right)<0}}1=\frac{1}{2t_{\varepsilon }}.\) 
It follows
\(\lim _{x\rightarrow \infty }\dfrac{\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })\gtrless 0\rbrace }{x}=\sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d\ne 0}}^{t_{\varepsilon }}\frac{1}{2 t_{\varepsilon }},\) 
and therefore
\(\lim _{x\rightarrow \infty }\dfrac{\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })\gtrless 0\rbrace }{\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })\ne 0\rbrace }=\frac{1}{2}.\) 
Case 2: \(\frac{\theta _p}{2\pi }=\frac{n}{m}\in (0,\frac{1}{2})\)  is rational, where \(m\)  and \(n\)  are coprime.
Write
\({\Re e\,}(a(p^{\nu })e^{-i\phi })=(\sin \theta _p)^{-1}\sin \left(\frac{2\pi (\nu +1) n}{m}\right)\cos \left(\frac{\pi j\nu }{r_{\varepsilon }}-\phi \right).\) 
By similar considerations as in the previous case, we have
\(\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })>0\rbrace =\sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d>0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(\frac{2\pi \nu n}{m}\right)>0}}1+ \sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d<0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(\frac{2\pi \nu n}{m}\right)<0}}1,\) 
and
\(\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })<0\rbrace =\sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d>0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(\frac{2\pi \nu n}{m}\right)<0}}1+ \sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d<0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(\frac{2\pi \nu n}{m}\right)>0}}1.\) 
Thus, the study of the distribution of signs of \({\Re e\,}(a(p^{\nu })e^{-i\phi })\) , turned out to study the distribution of signs of the sequence \(\left\lbrace \sin \left(\frac{2\pi \nu n}{m}\right)\right\rbrace _{\nu }\)  when \(\nu \)  runs through an arithmetic progression \(d\pmod {t_\varepsilon }\) . Note that this sequence is \(t_{\varepsilon }m^{\prime }\) -periodic and takes only finitely many different values, where \(m^{\prime }=\frac{m}{(m,t_\varepsilon )}\) . Thus, we can split this arithmetic progression into \(m^{\prime }\)  different sub-arithmetic progressions \(d+t_{\varepsilon }\ell \pmod {m^{\prime }t_\varepsilon }\)  where \(1\le \ell \le m^{\prime }\) , such that when \(\nu \)  runs through each of these sub-arithmetic progressions the sequence becomes constant and has always the same value, say \(s_{d,\ell }\) . Consequently
\(\nonumber \lim _{x\rightarrow \infty }\frac{1}{x}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(\frac{2\pi \nu n}{m}\right)>0}}1 &=& \lim _{x\rightarrow \infty }\frac{1}{x}\sum _{\genfrac{}{}{0.0pt}{}{\ell =1}{s_{d,\ell }>0}}^{m^{\prime }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\frac{\nu -d}{t_\varepsilon }\equiv \ell \!\!\!\!\!\pmod {m^{\prime }}}}1, \\&=& \lim _{x\rightarrow \infty }\frac{1}{x}\sum _{\genfrac{}{}{0.0pt}{}{\ell =1}{s_{d,\ell }>0}}^{m^{\prime }}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x}{\nu \equiv d+t_\varepsilon \ell \!\!\!\!\!\pmod {t_{\varepsilon }m^{\prime }}}}1,\\&=& \sum _{\genfrac{}{}{0.0pt}{}{\ell =1}{s_{d,\ell }>0}}^{m^{\prime }}\frac{1}{t_{\varepsilon }m^{\prime }}.\) 
Similarly, we obtain
\(\lim _{x\rightarrow \infty }\frac{1}{x}\sum _{\genfrac{}{}{0.0pt}{}{\nu \le x, \nu \equiv d \!\!\!\!\!\pmod {t_{\varepsilon }}}{\sin \left(\frac{2\pi \nu n}{m}\right)<0}}1=\sum _{\genfrac{}{}{0.0pt}{}{\ell =1}{s_{d,\ell }<0}}^{m^{\prime }}\frac{1}{t_{\varepsilon }m^{\prime }}.\) 
Therefore
\(\lim _{x\rightarrow \infty }\dfrac{\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })\gtrless 0\rbrace }{x}=\sum _{\genfrac{}{}{0.0pt}{}{d=1}{c_d\ne 0}}^{t_{\varepsilon }}\sum _{\genfrac{}{}{0.0pt}{}{\ell =1}{s_{d,\ell }\ne 0}}^{m^{\prime }}\frac{1}{2t_{\varepsilon }m^{\prime }},\) 
and hence
\(\lim _{x\rightarrow \infty }\dfrac{\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })\gtrless 0\rbrace }{\#\lbrace \nu \le x : {\Re e\,}(a(p^{\nu })e^{-i\phi })\ne 0\rbrace }=\frac{1}{2},\) 
as desired.

